Question: A curve in the plane is defined parametrically by the equations $x=6\tan(t)$ and $y=2t+1$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $3\sec^2(x)$ (Choice B) B $\dfrac{\cos^2(t)}{3}$ (Choice C) C $\dfrac{1}{3\cos^2(x)}$ (Choice D) D $\dfrac{\cot(t)}{3}$
In general, to find the derivative (i.e. the expression for $\dfrac{dy}{dx}$ ) of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ (where $u$ and $v$ are any functions of $t$ ), we use the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ We are given that $x=6\tan(t)$ and $y=2t+1$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\dfrac{d}{dt}\left(2t+1\right)}{\dfrac{d}{dt}(6\tan(t))} \\\\ &=\dfrac{2}{6\sec^2(t)} \\\\ &=\dfrac{\cos^2(t)}{3} \gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=\dfrac{\cos^2(t)}{3}$.